How do you evaluate the sum represented by $8 \sum n = 1 \frac{1}{n + 1}$ $sum_(n=1)^(8)1/(n+1)$ ?

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Answer 1 · 2014-10-22T23:25:25

Begin by changing the denominators to $1 + 1, 2 + 1, 3 + 1$ $1+1, 2+1, 3+1$ and so on to $8+1...$

Next add $1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9$

This requires a common denominator. If we multiply $9*8*7*5$ we will get $2520$ .

The $9$ picks up multiples of the $3$ , the and the $8$ picks up multiples of the $2, 3 and 4$ .

Now, multiply $1/2*1260/1260$ giving $1260/2520$ . Multiply $1/3*840/840$ giving $840/2520$ .

$1/4*630/630=630/2520, 1/5*504/504=504/2520, 1/6*420/420=420/2520, 1/7*360/360=360/2520, 1/8*315/315=315/2520 and 1/9*280/280=280/2520.$

Finally, add

$(1260+840+630+504+420+360+315+280)/2520$ which = $4609/2520$ .

How do you evaluate the sum represented by sum_(n=1)^(8)1/(n+1)8∑n=11n+1sum_(n=1)^(8)1/(n+1) ?